What volume of 0.320 M #KOH# is needed to react completely with 23.4 mL of 0.220 M #H_2SO_4#?

Answer 1

32.2mL

First we need to write out the balanced equation: #2KOH + H_2SO_4 → 2H_2O + K_2SO_4#

This indicates that two moles of KOH are required for every mole of sulfuric acid. (mL X M)2 = mL X M) using milliequivalents

0.00234L X 0.220mol/L = 0.000515 moles #H_2SO_4#

This will require 0.00103 moles of KOH based on our equation.

0.00322L, or 32.2mL, is equal to 0.00103mole/0.320 mol/L.

Verify: 10.304/2 = 5.148; 5.15 = 5.15; (32.2 x 0.320)/2 = 23.4 x 0.220

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Answer 2

Use the balanced equation: 2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O. Calculate moles of H₂SO₄, find the limiting reactant, and use its coefficient to determine moles of KOH needed. Finally, convert moles to volume using the given concentration of KOH. The required volume is 15.2 mL.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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